poj 3278 Catch That Cow (bfs搜索)
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Catch That Cow
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 46715 Accepted: 14673
Description
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Line 1: Two space-separated integers: N and K
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input
5 17
Sample Output
4
Hint
The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
开始看到这道题觉得和以前做的一道题很像,那道题直接用暴力做,一直判断是否可以整除2就行了,然后一开始也想用暴力去做,但是后来做的有些情况就考虑不到,这个题目也比那个题目复制 http://blog.csdn.net/whjkm/article/details/26985421 (以前那道题目的链接),后来仔细看了一下题,跟那道题还是有比较大的差别,那个题目简单就在于,它的起点是从0开始的,不能往后退;这个题目就有三种情况,向前进1,向后减1,使用传送向前进2倍的距离。开始也想不到要用bfs,看到了这个题目的分类是bfs,然后采用bfs来做。
这个题目的主要的思路就是 3 入口的bfs,先把第一个点的3种情况入队,标记这个点已经访问,再把下一个位置的3种情况入队,这里也有剪枝,(剪枝参考别人的)效率提高了不少,直到找到最终的那个点。
下面是代码,用数组模拟队列;
#include <cstdio>#include <cstring>const int maxn=200030;typedef struct Queue //一个结构体,保存这个点的位置,和到这个点所用的时间{ int count,step;}Queue;Queue queue[maxn];//队列数组int visit[maxn];//访问数组void bfs(int n,int k){ int front=0,rear=0; queue[rear].step=n;//把初始位置入队 queue[rear++].count=0; visit[n]=1;//标记访问 while(front<rear) { Queue q=queue[front++]; if(q.step==k)//找到终点位置 { printf("%d\n",q.count); break; } if(q.step-1>=0 && !visit[q.step-1])//注意剪枝的条件,一定要注意可以等于0 { visit[q.step]=1; queue[rear].step=q.step-1;//向后寻找一位 queue[rear++].count=q.count+1;//时间+1 } if(q.step<=k && !visit[q.step+1])//剪枝 { visit[q.step+1]=1; queue[rear].step=q.step+1;//向前寻找一位 queue[rear++].count=q.count+1; } if(q.step<=k && !visit[q.step*2])//剪枝 { visit[q.step*2]=1;//传送的情况 queue[rear].step=q.step*2; queue[rear++].count=q.count+1; } }}int main(){ int n,k; while(scanf("%d%d",&n,&k)!=EOF) { memset(visit,0,sizeof(visit)); bfs(n,k); } return 0;}
也可以用STL写,效率就没有自己手动写的高,还有标记数组用bool比较节省内存;
我自己用STL又写了一次,思路和上面的一样;
#include <cstdio>#include <cstring>#include <queue>using namespace std;const int maxn=200030;queue<int>q;int Count[maxn];bool visit[maxn];void bfs(int n,int k){ q.push(n); visit[n]=1; Count[n]=0; while(!q.empty()) { int x=q.front(); q.pop(); if(x==k) { printf("%d\n",Count[x]); break; } if(x-1>=0 && !visit[x-1]) { q.push(x-1); visit[x-1]=1; Count[x-1]=Count[x]+1; } if(x<=k && !visit[x+1]) { q.push(x+1); visit[x+1]=1; Count[x+1]=Count[x]+1; } if(x<=k && !visit[2*x]) { q.push(x*2); visit[x*2]=1; Count[x*2]=Count[x]+1; } }}int main(){ int n,k; while(scanf("%d%d",&n,&k)!=EOF) { memset(visit,0,sizeof(visit)); bfs(n,k); } return 0;}
#include <iostream>#include <queue>#define SIZE 100001using namespace std;queue<int> x;bool visited[SIZE];int step[SIZE];int bfs(int n, int k){int head, next;//起始节点入队x.push(n);//标记n已访问 visited[n] = true;//起始步数为0 step[n] = 0;//队列非空时 while (!x.empty()){//取出队头 head = x.front();//弹出队头 x.pop();//3个方向搜索 for (int i = 0; i < 3; i++){if (i == 0) next = head - 1;else if (i == 1) next = head + 1;else next = head * 2;//越界就不考虑了 if (next > SIZE || next < 0) continue;//判重 if (!visited[next]){//节点入队 x.push(next);//步数+1 step[next] = step[head] + 1;//标记节点已访问 visited[next] = true;}//找到退出 if (next == k) return step[next];}}}int main(){int n, k;cin >> n >> k;if (n >= k){cout << n - k << endl;}else{cout << bfs(n, k) << endl;}return 0;}
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